The argument for 1 is quite intricate. Over PID, a submodule of free module is free and ranks is strictly smaller than that of free module.
I will take $A\to B$ as embedding and $i:B\to A$ as a different injection. $F_X$ denotes free module surjective onto the generators of module $X$ and $K_X$ denotes the corresponding kernel.
$0\to K_A\to F_A\to A\to 0$
$0\to K_B\to F_B\to B\to 0$
$K_A$ is a submodule of free module and hence free. First use $A\to B$ to induce a morphism from $f_F:F_A\to F_B$ by projectivity. This induces morphism $f_K:K_A\to K_B$. From snake lemma, we have $0\to Ker(f_F)\to Ker(f_K)\to 0\to\dots$ where $\dots$ part I do not care. In particular, this says $Ker(f_F)$ is free and the morphism is restriction of $K_A\to F_A$. You can remove redundant part of $F_A$ and assume $f_F:F_A\to F_B$ and $f_K:K_A\to K_B$ injective. Now $f_F$ says $F_A$'s image is a submodule of $F_B$ which is free. In particular, the rank of $F_A$ must be strictly smaller than that of $F_B$. Now reverse argument by $i:B\to A$. You will see $F_A$'s rank is exactly that of $F_B$. Similarly for $K_A,K_B$. You can induce obvious morphism $K_A\to K_B$ and $F_A\to F_B$ by identity morphism as both $K_X,F_X$ are free for $X=A,B$. Thus $A\cong B$ from snake lemma again.
Unfortunately, my incapacity of figuring out a simpler argument. There should be some better argument to make it simpler proof.
The simpler proof of 1 comes from using chinese remainder theorem.
For 2, use PID module decomposition into torsion free part and torsion part. Then use chinese remainder theorem to see the rest.